Question

Vant hoff factor i for a dilute sol of the strong electrolyte bariumhydroxide is

Answer 1

Barium hydroxide is Ba(OH)2 . It dissociates: 
Ba(OH)2(aq) → Ba2+(aq) + 2OH-(aq) 
1mol Ba(OH)2 produces 3 mol ions 
van't Hoff factor = 3
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Answer 2

Answer:

Van't Hoff factor for barium hydroxide is 3.

Explanation:

Van't Hoff factor is used to calculate the extent of association or dissociation.

• Van't Hoff factor, i > 1 if there is a dissociation of the solute in the solution and i < 1 if there is an association of solute in the solution.
• For 100% dissociation of the solute, van't Hoff factor, i = number of ions produced from one molecule of the solute.

Van't Hoff factor, $i = \frac{Number of particles in solution}{Number of particles taken}$

Ba(OH)₂ undergoes complete dissociation as follows.

Ba(OH)₂₍aq₎ → Ba²⁺₍aq₎ + 2OH⁻₍aq₎

One mole of Ba(OH)₂ furnish one mole of Ba²⁺ions and two moles of OH⁻ ions i.e a total of 3 moles of ions in solution.

Therefore, for Ba(OH)₂

$i = \frac{3}{1} = 3$